泰州网站建设数字营销网站

题目描述

题解思路

我们只需要遍历长度长度为k的窗口，然后把窗口内数字之和填充到结果数组中的对应位置即可

题解代码

impl Solution {pub fn decrypt(code: Vec<i32>, k: i32) -> Vec<i32> {let n = code.len();let mut ans = vec![0i32; n];if k == 0 {return ans;}let mut sum = 0;if k > 0 {for i in 1..=k as usize {sum += code[i];}ans[0] = sum;for i in 1..n {sum += code[(i + k as usize) % n] - code[i];ans[i] = sum;}    } else {for i in k..0 {sum += code[n + i as usize];}ans[0] = sum;for i in 1..n {sum += code[i - 1] - code[(i + k as usize - 1 + n) % n];ans[i] = sum;}}ans}
}

题目链接

https://leetcode.cn/problems/defuse-the-bomb/